A plank fixed to a sled at rest in frame S, is of length ${\mathbf{L}}_{\mathbf{0}}$and makes an angle of ${\mathit{\theta }}_{\mathbf{0}}$with the x-axis. Later the sled zooms through frame S at a constant speed v parallel to the x-axis. Show that according to an observer who remains at rest in frame S, the length of the plank is now

$\mathbf{L}\mathbf{=}{\mathbf{L}}_{\mathbf{0}}\sqrt{\mathbf{1}\mathbf{-}\frac{{\mathbf{v}}^{\mathbf{2}}}{{\mathbf{c}}^{\mathbf{2}}}{\mathbf{cos\theta }}_{\mathbf{0}}}$

And the angle it makes with the x-axis is

$\mathit{\theta }\mathbf{=}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\left({\mathit{\gamma }}_{\mathit{v}}\tan {\mathit{\theta }}_0\right)$

Consider a plank fixed to a sled at rest in frame S, is of length ${\mathrm{L}}_0$and makes an angle of ${\theta }_0$with the x-axis.

Later Consider the sled zooms through frame S at a constant speed v parallel to the x-axis.

Write the formula of length of the plank.

$\mathbf{L}\mathbf{=}\sqrt{{\mathbf{L}}_{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{L}}_{\mathbf{y}}^{\mathbf{2}}}$ …… (1)

Here, ${\mathbf{L}}_{\mathbf{x}}$is length on x-axis and ${\mathbf{L}}_{\mathbf{y}}$is length on y-axis.

Write the formula of angle of the plank.

$\mathbf{tan}\left(\theta \right)\mathbf{=}\frac{{\mathbf{L}}_{\mathbf{y}}}{{\mathbf{L}}_{\mathbf{x}}}$ …… (2)

Here, ${\mathbf{L}}_{\mathbf{x}}$is length on x-axis and ${\mathbf{L}}_{\mathbf{y}}$is length on y-axis.

To understand why the Lorentz contraction is made along the direction of motion alone, I could advise going back and looking at issue number${}^{19}$. You'll see that as you go from one frame to the next, the perpendicular direction doesn't experience length contraction. So let's think about the contraction in the x-direction, knowing that the correct length will be the one determined in the plank's rest frame, which is the${}^{S^{\text{'}}}$frame (which is fixed to the plank).

The length on x-axis is:

$L_x^{\text{'}}=L_0\cos \left({\theta }_0\right)$

Since

$$\begin{gathered} L_x=\frac{L_x^{\text{'}}}{\gamma } \\ =\frac{L_0\cos \left({\theta }_0\right)}{\gamma } \end{gathered}$$

Determine the length of the plank.

Substitute$\frac{L_0\cos \left({\theta }_0\right)}{\lambda }$for$L_x^2$and$L_0\sin \left(\sin {\theta }_0\right)$for$L_y^2$into equation (1).

$$\begin{gathered} L=\sqrt{{\left(\frac{L_0\cos \left({\theta }_0\right)}{\gamma }\right)}^2+{\left(L_0\sin \left({\theta }_0\right)\right)}^2} \\ =L_0\sqrt{\left(1-\frac{v^2}{c^2}\right)+{\cos }^2\left({\theta }_0\right)+{\sin }^2\left({\theta }_0\right)} \end{gathered}$$

Since, ${\sin }^2\left({\theta }_0\right)+{\cos }^2\left({\theta }_0\right)=1$

Now,

$L=L_0\sqrt{1-\frac{v^2}{c^2}{\cos }^2\left({\theta }_0\right)}$

To make sure that this formula makes sense, you can look at the limiting situations, such as $\theta =0$or$\frac{\mathrm{\pi }}{2}$. One can anticipate no contraction at all if the plank is positioned at an angle of$\frac{\mathrm{\pi }}{2}$since the complete length is projected in the vertical direction. At an angle of 0, however, we still have the standard original Lorentz transformation.

Fig. 1

Now that we have a length contraction formula for a slanted plank, let's calculate how the angle will change as well. For frames travelling at a quicker rate, the horizontal distance is growing shorter while the vertical distance remains constant, therefore we would anticipate a larger angle as the relative velocity of the frames increases.

Determine the angle of the plank.

Substitute $L_0\sin \left({\theta }_0\right)$for $L_{\gamma }$and $\frac{L_0\cos \left({\theta }_0\right)}{\gamma }$for $L_x$into equation (2).

$$\begin{gathered} \tan \theta =\frac{L_0\sin \left({\theta }_0\right)}{{\displaystyle \frac{L_0\cos \left({\theta }_0\right)}{\gamma }}} \\ \theta ={\tan }^{-1}\left(\gamma \tan \left({\theta }_0\right)\right) \end{gathered}$$

Therefore, the value of angle of the plank is $\theta ={\tan }^{-1}\left(\gamma \tan \left({\theta }_0\right)\right)$.